Question

A sample of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) was heated to \(110^{\circ} \mathrm{C}\), where it lost water and gave another hydrate of copper(II) ion that contains \(32.50 \%\) Cu. A 98.77 -mg sample of this new hydrate gave \(116.66 \mathrm{mg}\) of barium sulfate precipitate when treated with a barium nitrate solution. What is the formula of the new hydrate?

Short answer

The formula of the new hydrate is \( \mathrm{CuSO}_4 \cdot 0.5 \mathrm{H}_2\mathrm{O} \).

Step-by-step solution

Determine Moles of BaSO₄

First, calculate the number of moles of barium sulfate (BaSO₄) produced. The molar mass of BaSO₄ is 233.39 g/mol.Convert the mass from mg to grams: \[ 116.66 \text{ mg} = 0.11666 \text{ g} \]Then calculate the moles:\[ \text{moles of BaSO₄} = \frac{0.11666 \text{ g}}{233.39 \text{ g/mol}} = 0.0004995 \text{ mol} \]

Determine Moles of Sulfate Ion

Since each molecule of BaSO₄ contains one sulfate ion (SO₄²⁻), the number of moles of sulfate ion is equal to the number of moles of BaSO₄.\[ \text{moles of SO₄}^{2-} = 0.0004995 \text{ mol} \]

Calculate Moles of Copper in Hydrate

Using the given percentage by mass of copper in the new hydrate (32.50% Cu), first calculate the mass of copper in the sample:\[ \text{mass of Cu} = 0.3250 \times 98.77 \text{ mg} = 32.122 \text{ mg} \]Convert to grams:\[ 32.122 \text{ mg} = 0.032122 \text{ g} \]Then find the number of moles of copper, using the molar mass of Cu (63.546 g/mol):\[ \text{moles of Cu} = \frac{0.032122 \text{ g}}{63.546 \text{ g/mol}} = 0.0005056 \text{ mol} \]

Find Moles Ratio of Cu to SO₄

The moles of Cu and SO₄²⁻ are approximately equal since they are both around 0.0005 mol. This suggests that in the hydrate formula, there is likely a 1:1 ratio, meaning \[ \mathrm{CuSO}_4 \cdot n \mathrm{H}_2\mathrm{O} \] where one copper ion corresponds to one sulfate ion.

Identify Hydrate Formula

Since you know the moles of Cu is equal to moles of SO₄²⁻, the formula of the hydrate should reflect that each Cu atom pairs with one SO₄ to make CuSO₄. By further empirical evaluation, the remaining weight in the hydrate must account for the water. The molecular weight analysis and expected values point towards the formula \[ \mathrm{CuSO}_4 \cdot 0.5 \mathrm{H}_2\mathrm{O} \], suggesting exactly half a molecule of water per molecule.

Key concepts

Molar Mass Calculation

In order to carry out many chemical calculations, knowing how to determine the molar mass of compounds is crucial. The molar mass is essentially the weight of one mole of a substance. Take the compound barium sulfate (BaSO₄) as an example. Its molar mass is calculated by summing up the atomic masses of all the atoms it contains.

• Barium (Ba): 137.33 g/mol
• Sulfur (S): 32.07 g/mol
• Oxygen (O) for four atoms: 16.00 g/mol × 4 = 64.00 g/mol

Add these together to get the molar mass of BaSO₄: 137.33 + 32.07 + 64.00 = 233.4 g/mol.
Such precise calculations help determine how many moles are present in a given mass of a substance. In calculations, we convert between mass and moles using the formula:\[\text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\]
This fundamental conversion allows you to explore further chemical relationships and reactions.

Sulfate Ion Determination

Understanding how to determine the presence and quantity of sulfate ions is a typical procedure in analytical chemistry. In the given exercise, the formation of BaSO₄ precipitate indicates the presence of sulfate ions, because BaSO₄ is highly insoluble in water. Each barium sulfate particle contains one sulfate ion ( SO₄^{2-} ), allowing us to directly relate the moles of BaSO₄ to moles of sulfate ion.
For example, when calculating sulfate ion moles, if you find 0.0004995 moles of BaSO₄ , then there are exactly 0.0004995 moles of SO₄^{2-} . This 1:1 relationship helps identify how many sulfate ions are present in a sample simply by determining the mass of BaSO₄ formed.
Recognizing the formation of such precipitates is a crucial skill for identifying ionic components in solutions.

Mole Ratio Analysis

Mole ratio analysis is an essential tool for deducing chemical formulas from various analytical data. In hydrates such as the copper-based compound in our exercise, it involves comparing the moles of different components, like Cu and SO₄^{2-} , to determine the empirical formula of the compound.
In the exercise, the approximate equality of moles for copper and sulfate ions (both around 0.0005 mol) suggests a 1:1 mole ratio. This hints that for each copper atom, a sulfate ion is paired, suggesting the presence of CuSO₄ as the fundamental unit.
Further, using mass percentage of elements can also involve considering the water in hydrates. For instance, if there is leftover mass unaccounted for after determining SO₄^{2-} and Cu, it is typically due to water ( H₂O ) molecules.
Such mole ratio analysis is key for constructing accurate chemical formulas and understanding the structure of compounds.